Question: What are the first three non-zero terms of the Maclaurin polynomial for the function $f(x)={{e}^{2x}}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $1\text{ }+\text{ }x\text{ }+\text{ }\frac{{{x}^{2}}}{2!}$ (Choice B) B $2\text{ }+\text{ 2}x\text{ }+\text{ }\frac{2{{x}^{2}}}{2!}$ (Choice C) C $2\text{ }+\text{ 4}x\text{ }+\text{ }\frac{8{{x}^{2}}}{2!}$ (Choice D) D $1\text{ }+\text{ 2}x\text{ }+\text{ }\frac{4{{x}^{2}}}{2!}$
Start with the Maclaurin series for $e^x$. $ e^x=1\text{ }+\text{ }x\text{ }+\text{ }\frac{{{x}^{2}}}{2!}\text{ }+\text{ }\frac{{{x}^{3}}}{3!}\text{ }+\ldots +\text{ }\frac{{{x}^{n}}}{n!}\text{ }+\ldots$ Substitute $2x$ for $x$. $ e^{2x}=1\text{ }+\text{ 2}x\text{ }+\text{ }\frac{4{{x}^{2}}}{2!}\text{ }+\text{ }\frac{8{{x}^{3}}}{3!}\text{ }+\ldots +\text{ }\frac{{{(2x)}^{n}}}{n!}\text{ }+\ldots $ Then the first three terms are $1~+~2x~+~\dfrac{4x^2}{2!}$.